
Noting that
\((M(f))^2(x)\le M(f^2)(x)\) 
due to Jensen's inequality. By (REF ), (REF ) and [1]} for \(\mathbb {B}=L^2(\lbrace 0<|z|\le 1\rbrace , \nu )\) , we get
\(\begin{split}&I_4^{(\delta )}(t)\\&=\int _0^t\int _{0<|z|\le 1}\left|g(X_{s-},z)-g(X^{(\delta )}_{s-},z)\right|^2\text{\rm {d}}t\nu (\text{\rm {d}}z)\\&\le 2C_2\int _0^t|Z_s^{(\delta )}|^2\Big \lbrace M\Big (\int _{0<|z|\le 1}|\nabla g(\cdot ,z)|^2\nu (\text{\rm {d}}z)\Big )(X_{s-})\\&\qquad \qquad \qquad \qquad \qquad +M\Big (\int _{0<|z|\le 1}|\nabla g(\cdot ,z)|^2\nu (\text{\rm {d}}z)\Big )(X^{(\delta )}_{s-})\Big \rbrace \text{\rm {d}}s\\&=2C_2\int _0^t|Z_s^{(\delta )}|^2\Big \lbrace M\Big (\int _{0<|z|\le 1}|U(\cdot ,z)|^2|z|^{2(\beta +\gamma -1)}\nu (\text{\rm {d}}z)\Big )(X_{s-})\\&\qquad \qquad \qquad \qquad \qquad +M\Big (\int _{0<|z|\le 1}|U(\cdot ,z)|^2|z|^{2(\beta +\gamma -1)}\nu (\text{\rm {d}}z)\Big )(X_{s-}^{(\delta )})\Big \rbrace \text{\rm {d}}s.\end{split}\) 
